Let $f(x)=\begin{cases} \dfrac{x+1}{\sqrt{x+5}-2}&\text{for }x\geq-5, x\neq -1 \\\\ k&\text{for }x=-1 \end{cases}$ $f$ is continuous for all $x>-5$. What is the value of $k$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-4$ (Choice B) B $0$ (Choice C) C $2$ (Choice D) D $4$
$\dfrac{x+1}{\sqrt{x+5}-2}$ is continuous for all $x>-5$ other than $x=-1$, which means $f$ is continuous for all $x>-5$ other than $x=-1$. In order for $f$ to also be continuous at $x=-1$, the following equality must hold: $\lim_{x\to -1}f(x)=f(-1)$ Since $f(-1)=k$, we will obtain the above equality by letting $k=\lim_{x\to -1}f(x)$. So let's find $\lim_{x\to -1}f(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to -1}f(x) \\\\ &=\lim_{x\to -1}\dfrac{x+1}{\sqrt{x+5}-2} \gray{\text{This is the rule for }x\neq-1} \\\\ &=\lim_{x\to -1}\dfrac{x+1}{\sqrt{x+5}-2}\cdot\dfrac{\sqrt{x+5}+2}{\sqrt{x+5}+2} \gray{\text{Rationalize}} \\\\ &=\lim_{x\to -1}\dfrac{(x+1)(\sqrt{x+5}+2)}{(x+5)-(2)^2} \gray{\text{Simplify}} \\\\ &=\lim_{x\to -1}\dfrac{\cancel{(x+1)}(\sqrt{x+5}+2)}{\cancel{x+1}} \gray{\text{Cancel common factors}} \\\\ &=\lim_{x\to -1}(\sqrt{x+5}+2) \\\\ &\text{(This is allowed because }x\neq -1) \\\\ &=\sqrt{(-1)+5}+2 \gray{\text{Direct substitution}} \\\\ &=4 \end{aligned}$ We obtained that if we set $k=4$, then $\lim_{x\to -1}f(x)=f(-1)$, which makes $f$ continuous at $x=-1$. Since we already saw that $f$ is continuous for any other $x>-5$, we can determine that it's continuous for all $x>-5$. In conclusion, $k=4$.